Is the Rotation Method Accurate for Drawing an Archimedean Spiral?

[person123]

I have a concern about a method for drawing an Archimedean spiral.

If a spiral is drawn by rotating a string around a rod, it would only be an approximation of an Archimedean spiral. Starting from the edge of the circle, each time the string goes around once, the change in distance from the center would be $\sqrt {r^2+(2πrx)^2}$ (using the Pythagorean Theorem) and the change in the angle would $2πx-\arctan(\frac{2πrx}{r})=2πx-\arctan(2πx)$ (x being the number of rotations around the rod). When dividing the two and plotting the graph, it doesn't show a perfectly horizontal line, which would be expected if it were truly Archimedean: https://www.desmos.com/calculator/v5c9rsbhd3

Do you think this approximation could be a problem? How would the accuracy of this method compare to plotting points and then connecting them free-handed? Thanks in advance.

 

[haruspex]

Do you think this approximation could be a problem?

It depends on the application.
The error is worst closest to the rod. At the extreme, the tangent to the curve goes through the centre of the rod.
I can visualise a better mechanism. The axis of the rod could be mounted on an axle which is itself on an arm length r (radius of rod) from a fixed point. Some gearing mechanism would ensure the rod is irrotational as its axis rotates about the fixed point. The string would be always tangential to the rod at the fixed point.

 

[mfb]

Do you think this approximation could be a problem?

Is there any application for high precision Archimedean spirals drawn manually?
It is not a problem if you just want to make a nice pattern on paper. If you want to produce parts that have to fit together, you'll use machines anyway, they will just follow an x-y-path given by a computer, they don't have that problem.

 

[person123]

Is there any application for high precision Archimedean spirals drawn manually?

Do you think drafting might require it (especially for architecture or something similar)? I also could see it being useful in school, maybe in a geometry class. I'm not entirely sure, though.

 

[person123]

I can visualise a better mechanism. The axis of the rod could be mounted on an axle which is itself on an arm length r (radius of rod) from a fixed point. Some gearing mechanism would ensure the rod is irrotational as its axis rotates about the fixed point. The string would be always tangential to the rod at the fixed point.

After starting this thread, I've thought of a way of drawing a true Archimedean spiral (I can't say I'm sure it will work, though).
Disc A is placed flat on the paper. Disc B is in contact with this one in this orientation. In this video, Disc A would be the wood disc and disc B would be one of the pink discs (ignore the differential):

Disc B is connected to a shaft and is allowed to slide across this shaft to change the gear ratio. This shaft is connected to a gear, and the gear is meshed with a rack. The edge of the rack would hold a pencil. Disc B, the shaft, the gear, and the rack, are allowed to rotate along the axis of rotation which goes through the center of Disc A. As it rotates, it causes Disc B to spin, causing the gear to spin, and causing the pencil to move outward or inward.
I'm not sure if I got the idea across clearly, but I think it's an interesting idea.

 

[haruspex]

This shaft is connected to a gear, and the gear is meshed with a rack.

You could simplify that a bit by using a threaded shaft, like a worm gear, and a corresponding thread in the hole of disk B.

 

[person123]

You could simplify that a bit by using a threaded shaft, like a worm gear, and a corresponding thread in the hole of disk B.

And the threaded shaft would be unable to rotate and instead be forced to move straight through while disc B rotates?

I think that may cause the system to get stuck and/or disc B to slip, especially if the threads were densely packed. But I may just not be imagining it correctly.

 

[haruspex]

the threaded shaft would be unable to rotate and instead be forced to move straight through while disc B rotates?

No, the shaft would be fixed to a vertical axis, unable to to rotate on its own axis and unable to migrate across the vertical axis. As it rotates about the vertical axis, the disk rolls on the ground, and so rotates on the shaft. That drives the disk towards/away from the vertical axis.

 

[person123]

I should have clarified that when I wrote it "would be unable to rotate", I meant rotating around its own axis. I'm assuming that disc B and the shaft are both rotating together around the vertical axis (causing the pencil to rotate around this axis as well).

This is quickly drifting away from math, so if we want to keep this going, I guess it should be moved.

 

[haruspex]

I should have clarified that when I wrote it "would be unable to rotate", I meant rotating around its own axis.

It was the shaft "moving straight through" that confused me. The disk moves along the shaft. Apart from rotating about the vertical axis the shaft does not go anywhere.

 

[person123]

Apart from rotating about the vertical axis the shaft does not go anywhere.

That drives the disk towards/away from the vertical axis.

I'm imagining disc B rotating around the vertical axis at a fixed distance from the center while the shaft (which is connected to the pencil) moves inward or outward while rotating around the vertical axis. This keeps the gearing ratio the same while drawing your spiral.

It seems like you're imagining that disc B is the one moving inward or outward.

 

[haruspex]

I'm imagining disc B rotating around the vertical axis at a fixed distance from the center while the shaft (which is connected to the pencil) moves inward or outward while rotating around the vertical axis. This keeps the gearing ratio the same while drawing your spiral.

It seems like you're imagining that disc B is the one moving inward or outward.

I was, with the pencil mounted adjacent to it. But it could work either way.

 

[person123]

Wouldn't moving disc B toward or away from the center change the gear ratio, making the spacing of the spiral uneven? (You would also need to prevent the pencil from rotating around the axis of disc B). If that weren't the case, moving the disc instead might be easier than moving the entire shaft.

 

[haruspex]

Wouldn't moving disc B toward or away from the center change the gear ratio

Ha! This approach cannot work. It will produce a geometric spiral.
As the shaft rotates dθ about the vertical axis the point of contact of the wheel with table moves rdθ (near enough). That will drive the wheel further away by αrdθ for some constant α. dr/dθ=αr.

 

[person123]

The camera quality is terrible, but this a rough orthographic drawing from top, front, and side as I would imagine it to look like. It uses a rack and pinion, the pencil is shown, and disc B is replaced by a ball (which I think would be more effective).